calculate the pH in a solution prepared by dissolving .10 mol of solid NH4Cl in .500 L of .40 M NH3. Assuming no volume change. Without the use of Henderson-hasselbalch

I posted the answer only a few minutes after you posted the restriction. How fast do you think I can type. Patience, please.

I see you gave me all of one minute before typing your same problem.

Ahh I'm so sorry! I only posted it again because I realized that I accidentally posted it in the answers section and not the question section.

That's the way you are supposed to do follow ups. I always check to see if there is a follow up. I think al of the volunteers do that although the time it takes us to get back varies according to how busy we are and how much time we have to spend answering questions. Hope you understood how to solve the problem. You might want to check and see if you get the same answer with this way vs the HH way. The HH way makes the same assumptions I suggested you make to keep from solving a quadratic.

To calculate the pH of the solution, we need to understand the equilibrium reactions involved in the system.

The solid NH4Cl dissolves in water to produce NH4+ and Cl- ions. Since Cl- is a spectator ion and does not affect the pH, our focus will be on NH4+.

NH4+ is a weak acid that can donate a proton (H+) to water. It can be represented as follows:

NH4+ + H2O ⇌ NH3 + H3O+

Now, let's determine the concentration of NH4+ ions in the solution.

Given that we dissolved 0.10 mol of NH4Cl in 0.5 L of solution, we can calculate the concentration of NH4+ ions:

Concentration of NH4+ = (0.10 mol) / (0.5 L) = 0.20 M

Next, we need to consider the dissociation of NH3 in water. NH3 is a weak base that can accept a proton (H+) from water. This reaction can be represented as:

NH3 + H2O ⇌ NH4+ + OH-

The initial concentration of NH3 is 0.40 M, and since it is in excess, we can assume that the concentration of NH4+ is the same.

Next, we need to determine the concentration of OH- ions. We can do this by applying the equilibrium constant expression for the above reaction:

Kw = [H3O+][OH-] = 1.0 x 10^-14 (at 25°C)

Since we assume no volume change, the concentration of [NH4+] will be equal to [NH3]. Therefore, we can consider x as the concentration of NH4+ and OH-, which will be the concentration of OH- generated.

Using the equilibrium constant expression for the reaction of NH3 and water, we get:

Kw = ([NH4+] + x)(x) = (0.40 + x)(x)

Since the concentration of OH- is x, the concentration of H3O+ will be (0.40 - x) due to the conservation of mass.

From the above reaction, we can write an expression for the equilibrium constant:

Kw = [H3O+][OH-] = (0.40 - x)(x) = 1.0 x 10^-14

Considering that x will be small compared to 0.40, we can approximate (0.40 - x) ≈ 0.40. So we have:

(0.40)(x) = 1.0 x 10^-14

Solving this equation, we find x ≈ 2.5 x 10^-14

Therefore, the concentration of OH- is 2.5 x 10^-14 M.

As the concentration of OH- increases, the concentration of H3O+ will decrease, which shifts the equilibrium of the NH4+ reaction towards the formation of more H3O+. Therefore:

[H3O+] ≈ 0.40 - x ≈ 0.40

Finally, we can calculate the pH using the expression:

pH = -log [H3O+]

Substituting the value of [H3O+], we get:

pH = -log(0.40)

Using a calculator, we find that the pH ≈ 0.4.

Therefore, the pH of the solution is approximately 0.4.